You can do it like this:

char data[sizeof(float)];


float f = 0.6f;

memcpy(data, &f, sizeof f);    // send data


float g;

memcpy(&g, data, sizeof g);    // receive data

In order for this to work, both machines need to use the same floating point representations.

As was rightly pointed out in the comments, you don't necessarily need to do the extra memcpy; instead, you can treat f directly as an array of characters (of any signedness). You still have to do memcpy on the receiving side, though, since you may not treat an arbitrary array of characters as a float! Example:

unsigned char const * const p = (unsigned char const *)&f;
for (size_t i = 0; i != sizeof f; ++i)
{
    printf("Byte %zu is %02X\n", i, p[i]);
    send_over_network(p[i]);
}

Excerpt from the C99 standard, normative annex F (The C++-standard does not explicitly mention this annex, though it includes all affected functions without change per reference. Also, the types have to match for compatibility.):

IEC 60559 floating-point arithmetic

F.1 Introduction

1 This annex specifies C language support for the IEC 60559 floating-point standard. The IEC 60559 floating-point standard is specifically Binary floating-point arithmetic for microprocessor systems, second edition (IEC 60559:1989), previously designated IEC 559:1989 and as IEEE Standard for Binary Floating-Point Arithmetic (ANSI/IEEE 754−1985). IEEE Standard for Radix-Independent Floating-Point Arithmetic (ANSI/IEEE 854−1987) generalizes the binary standard to remove dependencies on radix and word length. IEC 60559 generally refers to the floating-point standard, as in IEC 60559 operation, IEC 60559 format, etc. An implementation that defines __STDC_IEC_559__ shall conform to the specifications in this annex.356) Where a binding between the C language and IEC 60559 is indicated, the IEC 60559-specified behavior is adopted by reference, unless stated otherwise. Since negative and positive infinity are representable in IEC 60559 formats, all real numbers lie within the range of representable values.

So, include <math.h> (or in C++ maybe <cmath>), and test for __STDC_IEC_559__.

If the macro is defined, not only are the types better specified (float being 32bits and double being 64bits among others), but also the behavior of builtin operators and standard-functions is more specified.
Lack of the macro does not give any guarantees.

For x86 and x86_64 (amd64), you can rely on the types float and double being IEC-60559-conformant, though functions using them and operations on them might not be.

Easiest is to make a union:

#include <stdio.h>

int main(void) {
  int ii;
  union {
    float a;
    unsigned char bytes[4];
  } thing;

  thing.a = 1.234;
  for (ii=0; ii<4; ii++) 
    printf ("byte %d is %02x\n", ii, thing.bytes[ii]);
  return 0;
}

Output:

byte 0 is b6
byte 1 is f3
byte 2 is 9d
byte 3 is 3f

Note - there is no guarantee about the byte order… it depends on your machine architecture.

To get your function to work, do this:

void float2Bytes(byte bytes_temp[4],float float_variable){ 
  union {
    float a;
    unsigned char bytes[4];
  } thing;
  thing.a = float_variable;
  memcpy(bytes_temp, thing.bytes, 4);
}

Or to really hack it:

void float2Bytes(byte bytes_temp[4],float float_variable){ 
  memcpy(bytes_temp, (unsigned char*) (&float_variable), 4);
}

Note - in either case I make sure to copy the data to the location given as the input parameter. This is crucial, as local variables will not exist after you return (although you could declare them static, but let's not teach you bad habits. What if the function gets called again…)

You need to be sure the byte order is correct. Either

float f1 = *(float*)(received_data+1)
float f2 = *(float*)(received_data+5)

or if you have to reverse the bytes, you'll have to copy, so here's an alternative:

union {
  char chars[4];
  float f;
} u;

for (i = 0; i < 4; i++)
  u.chars[3-i] = received_data[i+1];
float f1 = u.f;     
// ... and similarly for second float

It had better be the case that these bytes were written by something like the inverse of this process (either you took the address of a float and cast it to char * or you had a union of char[4] and `float'.)

1)

Does sizeof(variable) simply get the variable type and return sizeof(type), which in this case would explain the result ?

Except for variable-length arrays, sizeof doesn't evaluate its operand. So yes, all it cares is the type. So sizeof(someFloatNumb) is 4 which is equivalent to sizeof(float). This explains printf("%i\n", sizeof(someFloatNumb));.

2)

[..] But I can still store a 8 bytes real value in a float, and my compiler says nothing about it. Does/Can gcc grow the capacity of a type ? (managing multiple variables behind the curtains to allow us that sort of things)

No. Capacity doesn't grow. You simply misunderstood how floats are represented/stored. sizeof(float) being 4 doesn't mean it can't store more than 2^32 (assuming 1 byte == 8 bits). See Floating point representation. What the maximum value of a float can represent is defined by the constant FLT_MAX (see <float.h>). sizeof(someFloatNumb) simply yields how many bytes the object (someFloatNumb) takes up in memory which isn't necessarily equal to the range of values it can represent. This explains why printf("%f\n", someFloatNumb); prints the value as expected (and there's no automatic "capacity growth").

3)

printf("%i\n", sizeof(281474976710656));

This is slightly more involved. As said before in (1), sizeof only cares about the type here. But the type of 281474976710656 is not necessarily int. The C standard defines the type of integer constants according to the smallest type that can represent the value. See https://stackoverflow.com/a/42115024/1275169 for an explanation.

On my system 281474976710656 can't be represented in an int and it's stored in a long int which is likely to be case on your system as well. So what you see is essentially equivalent to sizeof(long).

There's no portable way to determine the type of integer constants. But since you are using gcc, you could use a little trick with typeof:

typeof(281474976710656) x;
printf("%s", x); /* deliberately using '%s' to generate warning from gcc. */

generates:

warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘long int’ [-Wformat=] printf("%s", x);

P.S: sizeof results a size_t for which the correct format specifier is %zu. So that's what you should be using in your 1st and 3rd printf statements.

C standard does not say anything about the storage size of float and it has been purposely left out to the implementer.

Maybe on your system and compiler the size is 4. You can check that out by using sizeof(float). See also this discussion.